What is the sum from n=2 to infinity of 3/(n^3) ?

The sum from n=2 to infinity of 3/(n^3) is converges

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sum from n=2 to infinity of 3/(n^3)

n = 2 \sum \infty \frac{3}{n ^{3}}

converges

Solution steps

n = 2 \sum \infty \frac{3}{n ^{3}}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 3 \cdot n = 2 \sum \infty \frac{1}{n ^{3}}

Apply p-Series Test:

converges

= 3 converges

= converges

n = 7 \sum \infty \frac{1}{n} - \frac{1}{n + 1}

i = 1 \sum \infty (\frac{1}{2})^{i}

n = 0 \sum \infty \frac{1}{8 - n}

n = 1 \sum \infty 9 (0.1)^{n - 1}

n = 2 \sum \infty \frac{n ^{2}}{3 ^{n}}

What is the sum from n=2 to infinity of 3/(n^3) ?
The sum from n=2 to infinity of 3/(n^3) is converges